For a standard random variable Z let $\displaystyle \mu_n = E[Z^n]$

Show that:

$\displaystyle \mu_n = 0 $ when n is odd

$\displaystyle \mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j$

I started:

$\displaystyle E[e^tZ]=e^{{t^{2}}/2}$ from moment generating function of a normal variable

$\displaystyle = \displaystyle\sum_{j=0}^\infty \frac{(t^2/2)^j}{j!}$

$\displaystyle = \displaystyle\sum_{j=0}^\infty \frac{t^{2j}}{2^jj!}$

however I cannot see where to go now